This page discusses the unit impulse function and its Laplace Transform.
This function is also called the Delta Function and the Dirac Delta Function. It is usually represented mathematically as \( \delta(t) \).
Plot 1 - Unit Impulse Function |
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The unit impulse function is kind of a strange function. You can see the graph in plot 1. Even though we use an arrow in the graph, this is not a vector. The arrow represents the fact that the function height is infinity. We draw it from zero to one to represent that the area is one. Like the unit step function, this function can be used to filter another function. The unit impulse function filters a function at a specific point.
Formally, the unit impulse is a mathematical construct whose properties are that the area under the graph is one, with infinite height and zero width. Sounds impossible, right? Well, not in mathematics (math is cool, eh?)!
Unit Impulse Function | ||
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\(\displaystyle{ \delta (t) = \left\{ \begin{array}{lr} \infty & t = 0 \\ 0 & t \neq 0 \end{array} \right. }\) |
\(\displaystyle{ \int_{-\infty}^{+\infty}{\delta(t)~dt} = 1 }\) |
Here is a video explaining this in more detail.
video by Houston Math Prep |
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Since the area under the delta function is equal to one, it is easy to show that the Laplace Transform is equal to one.
\( \mathcal{L}\{ \delta(t) \} = 1 \) |
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Not as intuitively, when we shift in the time domain, the Laplace Transform is somewhat different. Here is an example. Try this on your own before looking at the solution.
Example
Calculate \( \mathcal{L}\{ \delta(t-a) \} \) where \( \delta(t) \) is the unit impulse function.
video by blackpenredpen |
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